Last Updated: 12 Jan, 2021
Allocate Books
Moderate
Asked in companies
Given an array ‘arr’ of integer numbers, ‘arr[i]’ represents the number of pages in the ‘i-th’ book.
There are ‘m’ number of students, and the task is to allocate all the books to the students.
Allocate books in such a way that:
1. Each student gets at least one book.
2. Each book should be allocated to only one student.
3. Book allocation should be in a contiguous manner.
You have to allocate the book to ‘m’ students such that the maximum number of pages assigned to a student is minimum.
If the allocation of books is not possible, return -1.
Example:
Input: ‘n’ = 4 ‘m’ = 2
‘arr’ = [12, 34, 67, 90]
Output: 113
Explanation: All possible ways to allocate the ‘4’ books to '2' students are:
12 | 34, 67, 90 - the sum of all the pages of books allocated to student 1 is ‘12’, and student two is ‘34+ 67+ 90 = 191’, so the maximum is ‘max(12, 191)= 191’.
12, 34 | 67, 90 - the sum of all the pages of books allocated to student 1 is ‘12+ 34 = 46’, and student two is ‘67+ 90 = 157’, so the maximum is ‘max(46, 157)= 157’.
12, 34, 67 | 90 - the sum of all the pages of books allocated to student 1 is ‘12+ 34 +67 = 113’, and student two is ‘90’, so the maximum is ‘max(113, 90)= 113’.
We are getting the minimum in the last case.
Hence answer is ‘113’.
Input format:
The first line contains two space-separated integers ‘n’ denoting the number of books and ‘m’ denotes the number of students.
The second line contains ‘n’ space-separated integers denoting the number of pages in each of ‘n’ books.
Output Format:
Return the integer as described above.
Note:
Do not print anything, just return the maximum number of pages that are assigned to a student is minimum.
Approaches
The basic idea is that, try each and every possible value that can be answered. It can be from ‘1’ to the sum of ‘arr’.
- Find the sum of all the elements of ‘arr’ in a variable ‘sum’.
- Iterate a loop ‘i’ from ‘1’ to ‘sum’(inclusive).
- Check for every ‘i’, if it is possible to divide the at least ‘i’ pages to every student.
- If possible then return ‘i’ because we are iterating from minimum then it is the minimum answer.
- Else check for ‘i+1’.
- If the number of ‘n<m’ then return ‘-1’, because the number of students is more than the number of books.
How to check - if it is possible to divide at least ‘mid’ number of pages in each student
Iterate a loop ‘i’, if sum of some contiguous books pages is less than or equals to mid then assign this sou to a single student and check for remain student and at the end if is possible then can say it is possible that ‘mid’ number of pages is possible to assign ‘m’ student.
The basic idea is use ‘binary search’, we check each and every value, if ‘x’ can answer then we try to minimise it, else check another value.
- Initially ‘min = 0’ and ‘max = sum of all pages’ and mid is ‘(max+min)/2’.
- Check if it is possible to divide at least ‘mid’ number of pages in each student.
- If possible then check for some small value ‘(min+mid)/2’.
- Else check some greater value ‘(mid+ max)/2’.
How to check - if it is possible to divide at least ‘mid’ number of pages in each student
Iterate a loop ‘i’, if sum of some contiguous books pages is less than or equals to mid then assign this sou to a single student and check for remain student and at the end if is possible then can say it is possible that ‘mid’ number of pages is possible to assign ‘m’ student.
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